# Complete metric space - sequences

The problem I'm trying is to prove whether or not the metric space of real-valued sequences $(x_n)$ such that $x_n=0$ for all but finitely many values of $n$, with the sup metric: $d((x_n),(y_n)) = max |x_n-y_n|$ is complete.

Now, I tried to find counterexamples but did not seem to find any...

Hence I'm trying to prove that if a sequence $((x_n)^{(k)}) : (x_n)^{(1)}, (x_n)^{(2)}, ...$ is Cauchy, then it converges to some sequence.

We can easily show how the sequence of the $k$th entries: $(x_k^{(n)})$ for a fixed $k$ is Cauchy in $\mathbb{R}$ and hence convergent. Moreover for all but finitely many values of $k$ we can find a subsequence of this series which is constantly zero. So for all but finitely many values of $k$, $(x_k^{(n)})\rightarrow 0$ by properties of real Cauchy sequences.

For finitely many values $k_1,k_2,...$, $(x_{k_{i}}^{(n)})\rightarrow \alpha_i$ where the limit could be nonzero.

This shows that $((x_n)^{(k)})\rightarrow (\alpha_n)$ where this limit sequence is still an element of the space we are considering, hence this space is complete.

This is my proof, but I feel the part in which I claim the existence of a constantly zero subsequence needs a little bit more care...

Any help or comment would be extremely helpful!!

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Let $(X,\|\cdot\|)$ denote the space you are describing.

For each $k\in\mathbb{N}$ define $Y_k$ to be the subspace of $X$, consisting of sequences $x=(x_j)$ such that $x_j=0$ for all $j>k$. It is easy to see each $Y_k$ is closed in $X$ and $X=\cup Y_k$.

However, $Y_k$ has empty interior. To see this, take $x=(x_j)\in Y_k$ and $\epsilon>0$. Then we can always define $y=(y_j)$ such that $y_j=x_j$ for $j\le k$, $y_{k+1}=\epsilon$ and $y_j=0$ for other $j$. Then $\|y-x\|<\epsilon$ but $y$ is not in $Y_k$.

Now each $Y_k$ is nowhere dense, so their countable union $X$ cannot be complete as a result of Baire's theorem.

Or, if you prefer concrete examples. Take $x^n$ to be the sequence whose first $n$ terms are $(1,\frac{1}{2},\dots,\frac{1}{n})$ and all other terms vanishes. You can see this $x^n$ is Cauchy in $X$, but you should feel they converge to the sequence $(1,\frac{1}{2},\frac{1}{3},\dots,\frac{1}{n},\frac{1}{n+1},\dots)$, which is not in $X$.

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thank you very much, I just started studying metric spaces so the first part of the answer is a bit obscure! –  Moritzplatz Nov 18 '12 at 14:29
so was the problem in my "proof" that part about choosing a subsequence? –  Moritzplatz Nov 18 '12 at 14:29
Yes, that part is a non sequitur. –  Hagen von Eitzen Nov 18 '12 at 14:31
thank you very much i got why now! –  Moritzplatz Nov 18 '12 at 14:32